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# ECA - Lecture 03

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### ECA - Lecture 03

1. 1. Electric Circuit Analysis Basic analysis techniques Ahsan Khawaja Ahsan_khawaja@comsats.edu.pk Lecturer Room 102 Department of Electrical Engineering
2. 2. Circuit Abstraction Question: What is the current through the bulb? Concept of Abstraction Solution: In order to calculate the current, we can replace the bulb with a resistor. R is the only subject of interest, which serves as an abstraction of the bulb.
3. 3. Voltage (Potential) a b VVab 5 a、b, which point’s potential is higher？ b a V6a V V4b V Vab = ? a b +Q from point b to point a get energy ，Point a is Positive? or negative ? Basic Quantities Example
4. 4. Voltage (Potential) ab c´ c d d´ 2211 21 221121222 2 21112 1111 111 1b1bb 0 )( )( , 0 rRrR EE I rRrRIEEIrEVIrVV EVV RrRIEIRVV rRIEIrVV IREVEV IRVIRVVVV V dda dd cd cc bc aab a Basic Quantities Example I
5. 5. Open Circuit R= I=0, V=E , P=0 E R0 Short Circuit R=0 E R0 R＝0 0 R E I 00 IREV 0 2 RIPE Basic Quantities
6. 6. Nodes, Branches, Loops, mesh • A circuit containing three nodes and five branches. • Node 1 is redrawn to look like two nodes; it is still one nodes. Kirchhoff's Current and Voltage Laws
7. 7. • sum of all currents entering a node is zero • sum of currents entering node is equal to sum of currents leaving node KCL KCL Mathematically i1(t) i2(t) i4(t) i5(t) i3(t) n j j ti 1 0)( n j jI 1 0 Kirchhoff's Current and Voltage Laws
8. 8. • sum of all currents entering a node is zero • sum of currents entering node is equal to sum of currents leaving node KCL P1.9 DCBA iiii Kirchhoff's Current and Voltage Laws In Out 0A B C O I I i i i i
9. 9. KCL + -120V 50* 1W Bulbs Is • Find currents through each light bulb: IB = 1W/120V = 8.3mA • Apply KCL to the top node: IS - 50IB = 0 • Solve for IS: IS = 50 IB = 417mA KCL-Christmas Lights Kirchhoff's Current and Voltage Laws
10. 10. KCL Current divider N V G1 G2 I + - I1 I2 121 21 11 1 11 RRR RR I R RI R V I I RR R I 21 1 2 Kirchhoff's Current and Voltage Laws In case of parallel : 1 2 1 2 1 1 1 , , V= I I G G G R R R R G
11. 11.  sum of voltages around any loop in a circuit is zero. KVL • A voltage encountered + to - is positive. • A voltage encountered - to + is negative. KVL Mathematically 0)( 1 n j j tv 0 1 n j jV Kirchhoff's Current and Voltage Laws
12. 12. KVL Determine the voltages Vae and Vec. Kirchhoff's Current and Voltage Laws 10 24 0ae V 16 12 4 6 0ae V 4 + 6 + Vec = 0
13. 13. KVL Voltage divider R1 R2 - V1 + + - V2 + - V 21 1 11 RR R VIRV 21 2 22 RR R VIRV Important voltage Divider equations N Kirchhoff's Current and Voltage Laws
14. 14. KVL Voltage divider kR 151 Volume control? Example: Vs = 9V, R1 = 90kΩ, R2 = 30kΩ Kirchhoff's Current and Voltage Laws
15. 15. Mesh Analysis • Technique to find voltage drops around a loop using the currents that flow within the loop, Kirchoff’s Voltage Law, and Ohm’s Law – First result is the calculation of the mesh currents • Which can be used to calculate the current flowing through each component – Second result is a calculation of the voltages across the components • Which can be used to calculate the voltage at the nodes.
16. 16. Mesh Analysis  What is a mesh:  It is a loop which does not contain any other loops within it. i1 i2 This circuit has two meshes
17. 17. Steps in Mesh Analysis Vin
18. 18. Step 1 • Identify all of the meshes in the circuit Vin
19. 19. Step 2 • Label the currents flowing in each mesh i1 i2 Vin
20. 20. Step 3 • Label the voltage across each component in the circuit i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 -
21. 21. Step 4 • Use Kirchoff’s Voltage Law i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 - 0 0 543 6321 VVV VVVVVin
22. 22. Step 5 • Use Ohm’s Law to relate the voltage drops across each component to the sum of the currents flowing through them. – Follow the sign convention on the resistor’s voltage. RIIV baR
23. 23. Step 5 i1 i2 + V1 _ Vin + V3 _ + V5 _ + V6 _ + V2 - + V4 - 616 525 424 3213 212 111 RiV RiV RiV RiiV RiV RiV
24. 24. Step 6 • Solve for the mesh currents, i1 and i2 – These currents are related to the currents found during the nodal analysis. 213 542 62171 iiI IIi IIIIi
25. 25. Step 7 • Once the mesh currents are known, calculate the voltage across all of the components.
26. 26. 12V
27. 27. From Previous Slides 616 525 424 3213 212 111 RiV RiV RiV RiiV RiV RiV 0 0 543 6321 VVV VVVVVin
28. 28. Substituting in Numbers kiV kiV kiV kiiV kiV kiV 1 3 6 5 8 4 16 25 24 213 12 11 0 012 543 6321 VVV VVVVV
29. 29. Substituting the results from Ohm’s Law into the KVL equations 0365 0158412 2221 12111 kikikii kikiikikiV
30. 30. Results • One or more of the mesh currents may have a negative sign. Mesh Currents ( A) i1 740 i2 264
31. 31. Results Currents ( A) IR1 = i1 740 IR2 = i1 740 IR3 = i1- i2 476 IR4 = i2 264 IR5 = i2 264 IR6 = i1 740 IVin = i1 740 The currents through each component in the circuit.
32. 32. Example Using Mesh analysis find the currents through each resistor KVL in loop 1: -2+2i1+4(i1-i2)= 0 6i1-4i2-2= 0 6i1-4i2= 2 …………..1) KVL in loop 2: 6+4(i2-i1)+i2= 0 -4i1+5i2= -6 ………..2) From eqn 1)&2) Loop currents: i1= -1A, i2= -2A Currents through each resistor: i2 = i1= -1A i1 = i2 = -2A i4 = i1-i2 = -1A-(-2A) = 1A i1 i2
33. 33. Example Use the mesh current method to determine the power associated with each voltage source in the circuit. Calculate the voltage vo across the 8 resistor. KVL in Loop 1: -40+2i1+8(i1-i2)=0 KVL in Loop 2: 8(i2-i1)+6i2+6(i2-i3)=0 KVL in Loop 3: 6(i3-i2)+4i3+20=0 i1 i2 i3
34. 34. 10i1-8i2+0i3=40 -8i1+20i2-6i3=0 0i1-6i2+10i3= -20 By solving the matrix i1=5.6A i2= 2A i3= -0.8A P40v= -40i1= -40(5.6) = -224W P20v = 20i3= 20(-0.8) = -16W vo= 8(i1-i2)= 8(5.6-2) = 8(3.6)=28.8V
35. 35. 35 Example Apply KVL to each mesh 2 1 7 5 0s V v v v 2 6 7 0v v v 15 3 0s v v v Mesh 1: Mesh 2: Mesh 3: 14 8 6 0s v v V vMesh 4: DC DC 1R 3R 5R 7R 2R 6R 8R 4R 1v 2v 3v 4v 5v 6v 7v 8v + + + + + + + + - - - - - - - - 1sV 2sV 1i 2i 3i 4i
36. 36. 36 DC DC 1R 3R 5R 7R 2R 6R 8R 4R 1v 2v 3v 4v 5v 6v 7v 8v + + + + + + + + - - - - - - - - 1sV 2sV 1i 2i 3i 4i 2 1 7 5 0s V v v v 2 6 7 0v v v 15 3 0s v v v Mesh 1: Mesh 2: Mesh 3: 14 8 6 0s v v V vMesh 4: 2 1 1 1 2 7 1 3 5 ( ) ( ) 0s V i R i i R i i R 2 2 2 4 6 2 1 7 ( ) ( ) 0i R i i R i i R 13 1 5 3 3 ( ) 0s i i R V i R Mesh 1: Mesh 2: Mesh 3: 14 4 4 8 4 2 6 ( ) 0s i R i R V i i RMesh 4: Express the voltage in terms of the mesh currents:
37. 37. 37 2 1 1 1 2 7 1 3 5 ( ) ( ) 0s V i R i i R i i R 2 2 2 4 6 2 1 7 ( ) ( ) 0i R i i R i i R 13 1 5 3 3 ( ) 0s i i R V i R Mesh 1: Mesh 2: Mesh 3: 14 4 4 8 4 2 6 ( ) 0s i R i R V i i RMesh 4: Mesh 1: Mesh 2: Mesh 3: Mesh 4: 21 5 7 1 7 2 5 3 ( ) s R R R i R i R i V 7 1 2 6 7 2 6 4 ( ) 0R i R R R i R i 15 1 3 5 3 ( ) s R i R R i V 16 2 4 6 8 4 ( ) s R i R R R i V
38. 38. 38 Mesh 1: Mesh 2: Mesh 3: Mesh 4: 21 5 7 1 7 2 5 3 ( ) s R R R i R i R i V 7 1 2 6 7 2 6 4 ( ) 0R i R R R i R i 15 1 3 5 3 ( ) s R i R R i V 16 2 4 6 8 4 ( ) s R i R R R i V 2 1 1 1 5 7 7 5 1 7 2 6 7 6 2 5 3 5 3 6 4 6 8 4 0 00 0 0 0 0 s s s VR R R R R i R R R R R i VR R R i R R R R i V